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In a Young's double slit experiment slit...

In a Young's double slit experiment slits are separated by 0.5mm and the screen is placed 150 cm away. A beam of light consisting of two wavelength 650 nm and 520 nm is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright due to both the wavelengths coincide is

A

1.56 mm

B

7.8 mm

C

9.75 mm

D

15.6 mm

Text Solution

Verified by Experts

Let the m th bright fringe due to the wavelength 650 nm superposes with the nth bright fringe due to wavelength 520 nm.
Then , `(mlambda_(1)D)/(d)=(nlambda_(2)D)/(d)`
or `(m)/(n)=(lambda_(2))/(lambda_(1))=(520)/(650)=(4)/(5)`
In case of least distance,
m=4 and n = 5
`:.` the least distance `=(m lambda D)/(d)`
`=(4xx(650 xx10^(-7))xx150)/(0.5 xx 10^(-1))`cm
=7.8 mm
The option B is correct.
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